Prove function $f:\mathbb{R} \rightarrow \mathbb{R}$, $f(x)=e^{x^2}$ is not uniformly continuous. I try this proof by contradiction that
Assuming $f:\mathbb{R} \rightarrow \mathbb{R}$, $f(x)=e^{x^2}$ is uniformly continuous.
By definition we have $\forall \varepsilon>0$ $\exists \delta>0$ $|x-y|<\delta \Rightarrow |f(x)-f(y)|<\varepsilon$.
Choosing $\varepsilon=1$,we have $|x-y|<\delta \Rightarrow |f(x)-f(y)|<1$.
If $x<y$, than $|f(x)-f(y)|=|e^{y^2}-e^{x^2}|<|e^{(x+\delta)^2}-e^{x^2}|=|e^{x^2}(e^{\delta^2+2x\delta}-1)|<1$.
But for $x \rightarrow +\infty$, we have $|e^{x^2}(e^{\delta^2+2x\delta}-1)| \rightarrow +\infty$, which is a contradiction.
For $x>y$ it is the same.
May I ask if this is reasoning correct and sufficient?
Many thanks!
 A: I am just a student so i hope my answer is correct. But maybe you can try this.
The definition of a no uniformly continuous function is: $\exists \epsilon_0>0, \forall \delta>0 \; s.t. \; \exists |x-y|<\delta \Rightarrow |f(x)-f(y)|\geq \epsilon_0$
So if we find two sequences $x_n,y_n$ that take value on the definition domain of the function $f(x)$ and moreover that verify $\lim_{n\rightarrow \infty } |x_n-y_n| =0$ and  $lim_{n\rightarrow \infty }|f(x_n)-f(y_n)|=\infty$. It simply means that it exist $x,y$ that will belong to the set: $S_{\delta}=\left \{ x,y \in definition \; domain \; of \; the \; fct \; f : |x-y|<\delta \right \}$ for any $\delta>0$ you can choose. But on the other side we have: $|f(x_n)-f(y_n)|>\epsilon_0$ for any $\epsilon_0>0$ you can choose from a specific $N$ (definition of divergence to $+\infty$).
Following this logic, let choose: $x_n=n+1/n \; , \; y_n=n \Rightarrow \lim_{n\rightarrow \infty } |x_n-y_n| = \lim_{n\rightarrow \infty } |1/n|=0 $ but on the other side: $ lim_{n\rightarrow \infty }|e^{(n+1/n)^2}-e^{(n)^2}|=\infty $
Indeed: $|e^{(n+\frac{1}{n})^2}-e^{n^2}|=|e^{n^2+\frac{1}{n^2}+2}-e^{n^2}|=|e^{n^2}(e^{\frac{1}{n^2}+2}-1)|\underset{n \to \infty }{\rightarrow} \infty |(e^2-1)|=\infty$
